The example goes as follows
The set X = {1,2} × $\mathbb{Z^+}$ in the dictionary order is another example of an ordered set with a smallest element. Denoting $1×n$ by $a_n$ and $2×n$ by $b_n$ We can represent X by $a_1,a_2,..$;$b_1,b_2,...$ The order topology on X is not a discrete topology. Most one point sets are open but there is an exception — the set {$b_1$}. Any open set containing $b_1$ must contain a basis element about $b_1$ and any basis element containing $b_1$ contains the points of the $a_i$ sequence
I have no idea how to prove the text in bold so as to understand why does the basis element have to contain the points of the $a_i$ sequence any help would be appreciated
Sincerely, Aditya
Recall that basis elements for the order topology are the open intervals, so in this case, for $b_1$ to be in $(r,s)$, with $r,s \in X$, $r < s$, it is necessary (by definition) that $r < b_1$, so that $r = a_n$ for some $n \in \mathbb N$, as well as $s > b_1$, so that $s = b_k$ for some $k \in \mathbb N$ as well. But this implies, for example, that $a_{n+1} \in (r,s) = (a_n,b_k)$ as the author claims.
Note that the semi-open intervals of the form $[a_1, x)$ are also open here, but this case is not problematic.